Beer bottles are filled so that they contain an average of 4

Beer bottles are filled so that they contain an average of 415 ml of beer in each bottle. Suppose that the amount of beer in a bottle is normally distributed with a standard deviation of 6 ml. Use Table 1.

What is the probability that a randomly selected bottle will have less than 410 ml of beer?(Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)

What is the probability that a randomly selected 8-pack of beer will have a mean amount less than 410 ml? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)

What is the probability that a randomly selected 15-pack of beer will have a mean amount less than 410 ml? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)

Solution

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    410      
u = mean =    415      
          
s = standard deviation =    6      
          
Thus,          
          
z = (x - u) / s =    -0.83      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.83   ) =    0.203269392 [answer]

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    410      
u = mean =    415      
n = sample size =    8      
s = standard deviation =    6      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -2.36      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.36   ) =    0.009137468 [ANSWER]
          

C)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    410      
u = mean =    415      
n = sample size =    15      
s = standard deviation =    6      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -3.23      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -3.23   ) =    0.000618951 [ANSWER]